Lattices from equiangular tight frames Albrecht B¨ottcher, Lenny Fukshansky, Stephan Ramon Garcia, Hiren Maharaj, Deanna Needell Abstract. We consider the set of all linear combinations with integer coefficients of the vectors of a unit equiangular tight (k, n) frame and are interested in the question whether this set is a lattice, that is, a discrete additive subgroup of the k-dimensional Euclidean space. We show that this is not the case if the cosine of the angle of the frame is irrational. We also prove that the set is a lattice for n = k + 1 and that there are infinitely many k such that a lattice emerges for n = 2k. We dispose of all cases in dimensions k at most 9. In particular, we show that a (7,28) frame generates a strongly eutactic lattice and give an alternative proof of Roland Bacher’s recent observation that this lattice is perfect. AMS classification. Primary: 15B35, Secondary: 05B30, 11H06, 42C15, 52C07. Keywords. Lattice, Equiangular lines, Tight frame, Conference matrix.
1
Introduction
Let 2 ≤ k < n and let G be a real k × n matrix. Denote the columns of G by f1 , . . . , fn . These columns or G itself are called a unit equiangular tight (k, n) frame if GG0 = γI p with γ = n/k (tightness) and G0 G = I + (1/α)C with α = k(n − 1)/(n − k) and a matrix C whose diagonal entries are zero and the other entries of which are ±1 (property of being equiangular unit vectors). Define Λ(G) = spanZ {f1 , . . . , fn }. Our investigation is motivated by the following question. When is Λ(G) a lattice, that is, a discrete additive subgroup of Rk ? In case it is a lattice, what are its geometric properties? After having posed the question in its most concise form, some comments are in order. By Rk we understand the column-wise written Euclidean Rk with the usual scalar product (·, ·). The condition G0 G = I+(1/α)C with C as above means that kfj k = 1 for all j and that |(fi , fj )| = 1/α for i 6= j. In other words, the vectors fj are all unit vectors and each pair of them makes the angle ϕ or π−φ such that | cos ϕ| = | cos(π−φ)| = 1/α. The equality GG0 = γI is equivalent to the requirement that kG0 xk2 = γkxk2 for all x P in Rk , which in turn is the same as saying that nj=1 (fj , x)2 = γkxk2 for all x ∈ Rk . It is well known since [23, 24] that the two equalities G0 G = pI + (1/α)C with C as above and GG0 = γI necessarily imply that γ = n/k and α = k(n − 1)/(n − k). The set Λ(G) is always a group. It is the discreteness requirement that may prevent Λ(G) from being a lattice. Consider, for example, the unit equiangular tight (3, 6)
1
frame G that is induced √ by the 6 upper vertices of a regular icosahedron. As shown in [24], with p = (1 + 5)/2, this frame is given by the columns of the matrix 0 0 1 −1 p p 1 1 −1 p p 0 0 . G= p 1 + p2 p p 0 0 1 −1 By Dirichlet’s approximation theorem, for which see, e.g., p. 17 of [21], there are integers xn , yn such that yn → ∞ and xn + p ≤ 1 . y2 yn n In particular, xn + pyn → 0 as n → ∞. The linear combination of the columns of G with the coefficients xn + yn , yn − xn , yn , yn , xn , −xn , equals 0 1 2(xn + yn p) , p 1 + p2 2(yn p + xn ) 3 which tends to zero as n → ∞. Consequently, Λ(G) is not √ a discrete subgroup of R and thus it is not a lattice. Note that in this case α is 5 andpthus irrational. We will show in Proposition 2.1 that Λ(G) is never a lattice if α = k(n − 1)/(n − k) is irrational.
Equiangular tight frames (ETFs) possess many properties similar to orthonormal bases, yet may also be highly overcomplete, making them very attractive in many applications. For this reason there has been a recent surge of work addressing the construction and analysis of these frames. They appear in many practical applications, such as error correcting codes [12, 23], wireless communications [22, 23], security [15], and sparse approximation [9, 18, 25, 26]. In sparse approximation for example, the incoherence (small 1/α, the absolute value of pairwise inner products of vectors) of ETFs allows them to be used as sensing operators. Viewing an ETF as a matrix whose columns consist of the frame vectors, samples of a signal are acquired via inner products between the signal and the rows of this (typically highly underdetermined) matrix. Under the assumption that the signal vector is sparse (has a small number of nonzero coordinates), the signal can be accurately reconstructed from this compressed representation. However, in many applications there is more known about the signal than it simply being sparse. For example, in error correcting codes [5] and communications applications like MIMO [17] and cognitive radio [1], the signal vectors may come from some lattice. However, there has been very little rigorous mathematical developments on the intersection between arbitrary lattice-valued signals and sparse approximation (see e.g. [10] and references therein). In this work, we attempt to take the first step toward a rigorous analysis of properties of equiangular tight frames and associated lattices. We are especially interested in the 2
following questions. When does the integer span of an ETF form a lattice? Does this lattice have a basis of minimal vectors? Is the generating frame contained among the minimal vectors of this lattice? We also study further geometric properties of the resulting lattices, such as eutaxy and perfection. Our hope is that this investigation will contribute not only to the understanding of ETFs in general, but also to their explicit use in applications with lattice-valued signals. For example, if the integer span of an ETF is a lattice, then the image of that ETF viewed as a sensing matrix restricted to integer-valued signals forms a discrete set. In some sense this is analogous to the well-known Johnson-Lindenstrauss lemma [3], and may be used to provide reconstruction guarantees for ETF sampled signals. More concretely, if the lattice constructed from the ETF is such that its minimal vectors are the frame vectors themselves, this guarantees a minimum separation between sample vectors in its image. These types of properties are essential for sparse reconstruction and can be leveraged to design new sampling mechanisms and reconstruction guarantees. On the other hand, it is also useful to know when such properties are impossible. We leave a detailed analysis and link to applications as future work, and focus here on the mathematical underpinnings to the questions raised above.
2
Main results
Let L be a lattice in Rk , and let V = spanR L be the subspace of Rk that it spans. Then the rank of L, denoted by rk(L), is defined to be the dimension of V . We say that L has full rank if V = Rk . The minimal distance of a lattice L ⊂ Rk is defined as d(L) = min{kxk : x ∈ L \ {0}}. (Note that if L is only required to be a subgroup of Rk , then L is a lattice if and only if inf{kxk : x ∈ L \ {0}} > 0.) The set of minimal vectors, S(L), is the set of all x ∈ L with kxk = d(L). The lattice L is called well-rounded if Rk = spanR S(L), and we say that it is generated by its minimal vectors if L = spanZ S(L). It follows from a well-known theorem of van der Waerden [27] (see also §§2.2-2.3 of [19]) that the second condition is strictly stronger than the first when rk(L) ≥ 5. Furthermore, Conway and Sloane [7] demonstrated that when rk(L) ≥ 11 the lattice L may be generated by minimal vectors, but not contain a basis of minimal vectors, i.e., while L = spanZ S(L) there may not exist Rlinearly independent vectors f1 , . . . , frk(L) ∈ S(L) such that spanZ {f1 , . . . , frk(L) } = L; Martinet and Sch¨ urmann [14] further showed that this is possible when rk(L) ≥ 10, but cannot happen when rk(L) ≤ 9. A finite subset {q1 , . . . , qm } of the unit sphere Σk−1 in Rk is called a spherical t-design for a positive integer t if for every real polynomial p of degree ≤ t in k variables, k
Z p(x) dσ(x) = Σk−1
1 X p(qi ), m i=1
where dσ denotes the unit normalized surface measure on the sphere Σk−1 . A full rank lattice in Rk is called strongly eutactic if its set of minimal vectors (normalized to lie 3
on Σk−1 ) forms a spherical 2-design. We finally define the notion of a perfect lattice. Recall that we write vectors x in Rk as column vectors. A full rank lattice L in Rk is called perfect if the set of symmetric k × k matrices {xx0 : x ∈ S(L)} spans all real symmetric k × k matrices as an R-vector space. Two lattices L and M in Rk are called similar if L = aU M for some a ∈ R and some orthogonal k × k matrix U . Conditions such as well-roundedness, generation by minimal vectors, existence of bases of minimal vectors, strong eutaxy, and perfection are preserved on similarity classes of lattices. Furthermore, there are only finitely many strongly eutactic and only finitely many perfect similarity classes of lattices in Rk for each k ≥ 1. Given a full rank lattice L ⊂ Rk , it is possible to associate a sphere packing to it by taking spheres of radius d(L)/2 centered at every point of L. It is clear that no two such spheres will intersect in their interiors. Such sphere packings are usually called lattice packings. One convenient way of thinking of a lattice packing is as follows. The Voronoi cell of L is defined to be V(L) := {x ∈ Rk : kxk ≤ kx − yk ∀ y ∈ L}. Then Rk is tiled with translates of the Voronoi cell by points of the lattice, and spheres in the packing associated to L are precisely the spheres inscribed in these translated Voronoi cells. A compact measurable subset of Rk is called a fundamental domain for a lattice L if it is a complete set of coset representatives in the quotient group Rk /L. All fundamental domains of the same lattice have the same volume, and the Voronoi cell of a lattice is an important example of a fundamental domain. A central problem of lattice theory is to find a lattice in each dimension k ≥ 1 that maximizes the density of the associated lattice packing. There is an easy formula for the packing density of a lattice. A lattice L in Rk can be written as L = BZk , where B is a basis matrix of L, i.e., the columns of B form a basis for L. The determinant p of L is then defined to be det L := det(B 0 B), which is an invariant of the lattice, since any two basis matrices of L are related by a integer linear transformation with determinant ±1. The significance of the determinant is given by the fact that it is equal to the volumes of the fundamental domains. It is then easy to observe that the density of the lattice packing associated to L is the volume of one sphere divided by the volume of the translated Voronoi cell that it is inscribed into, that is, δ(L) :=
ωk d(L)k , 2k det L
(1)
where ωk is the volume of the unit ball in Rk . In fact, this packing density function δ is defined on similarity classes of lattices in a given dimension, and a great deal of attention in lattice theory is devoted to studying its properties. There is a natural quotient metric topology on the space of all full rank lattices in Rk , given by identifying this space with GLk (R)/ GLk (Z): indeed, every A ∈ GLk (R) is a basis matrix of some lattice, and A, B ∈ GLk (R) are basis matrices for the same lattice if and only if A = U B 4
for some U ∈ GLk (Z). A lattice is called extreme if it is a local maximum of the packing density function in its dimension: this is a particularly important class of lattices that are actively studied. A classical result of Voronoi states that perfect strongly eutactic lattices are extreme (see, for instance, Theorem 4 of [20]); on the other hand, if a lattice is strongly eutactic, but not perfect, then it is a local minimum of the packing density function (see Theorem 9.4.1 of [13]). Good sources for further information about lattice theory, the sphere packing problem, and the related arithmetic theory of quadratic forms are Martinet’s book [13], the well-known book by Conway and Sloane [8], and the more recent book by A. Sch¨ urmann [19]. We now return to our construction Λ(G) from unit equiangular frames and describe our results. It is well known that unit equiangular tight (k, k + 1) frames exist for all k ≥ 2. According to [24], except for the (k, k + 1)-case, the only unit equiangular tight (k, n) frames with k ≤ 9 are (3, 6), (5, 10), (6, 16), (7, 14), (7, 28), (9, 18)
(2)
frames. Our first result says the following. p Proposition 2.1 If Λ(G) is a lattice, then α = k(n − 1)/(n − k) must be a rational number. √ √ Thus, since α = 1/ 5 for the (3, 6) frame, α = 1/ 13 for the (7, 14) frame, and √ α = 1/ 17 for the (9, 18) frame, these three frames do not generate lattices. (The fact that the (3, 6) frame does not induce a lattice was proved in Section 1 by a different argument.) We will show that there are unit equiangular tight (5, 10), (6, 16), and (7, 28) frames which generate lattices. Moreover, we will prove the following results. Theorem 2.2 (a) For every k ≥ 2, there are unit equiangular tight (k, k +1) frames G such that Λ(G) is a full rank lattice. The lattice Λ(G) has a basis of minimal vectors, it is non-perfect and strongly eutactic, and hence it is a local minimum of the packing density function in dimension k. (b) There are infinitely many k for which there exist unit equiangular tight (k, 2k) frames G such that Λ(G) is a full rank lattice. (c) There is a unit equiangular tight (7, 28) frame G for which Λ(G) has a basis of minimal vectors, is a perfect strongly eutactic lattice, and hence extreme. Remark 2.3 We explicitly construct the lattices of Theorem 2.2. We show that those of parts (a) and (c) and those with k ≤ 13 of part (b) have the property that the set of minimal vectors consists precisely of ± the generating frame vectors. The well known result of Gerzon (see, for instance, Theorem C of [24]) asserts that for a (k, n) equiangular tight frame necessarily n ≤ k(k + 1)/2. On the other hand, k(k + 1)/2 is the minimal number of (± pairs of) minimal vectors necessary (but not sufficient) for a lattice in Rk to be perfect. Since only very few equiangular tight frames achieve equality in Gerzon’s bound, it is likely quite rare for perfect lattices to be generated 5
by equiangular tight frames. Perfection is a necessary condition for extremality, and hence it is unreasonable to expect to obtain extreme lattices often in this way. The only such example we have discovered is the lattice from the (7, 28) frame in part (c) of our Theorem 2.2, perfection of which has also previously been discussed in [2]. The strong eutaxy of our lattice constructions in Theorem 2.2(a),(c) is established directly with the use of the following result. Proposition 2.4 Suppose that Λ(G) is a lattice and S(Λ(G)) = {±f1 , . . . , ±fn }. Then Λ(G) is strongly eutactic. Proof. A spanning set {g1 , . . . , gm } for Rk is called a Parseval frame if kxk2 = Pm 2 k j=1 (gj , x) for all x ∈ R . Further, {g1 , . . . , gm } is a spherical 2-design if and only if o np p k/m g1 , . . . , k/m gm is a Parseval frame and
Pm
i=1
gi = 0 (see [12] for details, especially Proposition 1.2).
Now let G = (f1 . . . fn ) be a unit equiangular tight (k, n) frame, and assume that Λ(G) is a lattice such that S(Λ(G)) = {±f1 , . . . , ±fn }. We then have kxk
2
n k X = (x, fj )2 + (x, −fj )2 2n j=1 !2 !2 r r n X k k x, fj + x, − fj , = 2n 2n j=1
n p o p for every x ∈ Rk . Hence ± k/2n f1 , . . . , ± k/2n fn is a Parseval frame, and therefore S(Λ(G)) is a spherical 2-design. A summary of a part of our results is given in Table 1.
3
Rationality of the cosine of the frame
Suppose G is a unit tight (k, n) frame. Then GG0 = γI and hence G has rank k. Let G0 be the k × k matrix formed by arbitrarily chosen k linearly independent columns of G and denote by G1 the k × (n − k) matrix constituted by the remaining columns. We may without loss of generality assume that G = (G0 G1 ). We emphasize that G0 is invertible. Recall that Λ(G) is called a full-rank lattice if spanR {f1 , . . . , fn } is all of Rk . Note that in the following proposition we do not require equiangularity. Proposition 3.1 Let G = (G0 G1 ) be a unit tight (k, n) frame. Then the following are equivalent. (i) Λ(G) is a lattice. 6
Table 1: Summary of a part of our results.
(k, n)
cosine
1 α
1 (k + 1, k) k
Volume of a
S(Λ) = {±f1 , . . . , ±fn }?
fundamental domain
Basis of minimal vectors?
1 √ k+1
1 1+ k
k/2 Yes, Yes
(3, 6)
1 √ = 0.4472 5
no lattice
(5, 10)
1 3
4 = 0.4444 9
Yes, Yes
(6, 16)
1 3
23 = 0.2963 33
Yes, Yes
(7, 14)
1 √ = 0.2774 no lattice 13
(7, 28)
1 3
(9, 18)
1 √ = 0.2425 no lattice 17
(13, 26)
1 5
26 = 0.0458 59/2
(25, 50)
1 7
211 · 3 · 5 · 112 = 0.00071052 ?, ? 723/2
23 = 0.1711 37/2
7
Yes, Yes, and perfect
Yes, Yes
(ii) Λ(G) is a full rank lattice. (iii) There exist β ∈ Z \ {0} and X ∈ Zk×(n−k) such that G−1 0 G1 = (1/β)X. If (iii) holds with β = 1, then G0 is a basis matrix for Λ(G). Proof. Since G0 is invertible, we have spanR {f1 , . . . , fn } = Rk , which proves the equivalence of (i) and (ii). Suppose (ii) holds. Then G0 = BX0 and G1 = BX1 with an invertible k × k matrix B and integer matrices X0 , X1 . The matrix X0 is invertible, so B = G0 X0−1 and hence G1 = G0 X0−1 X1 = G0
1 1 X2 X1 = G0 X det X0 β
with β = det X0 and X = X2 X1 . This proves (iii). Conversely, suppose (iii) is true. It is clear that Λ(G) = spanZ {f1 , . . . , fn } is an additive subgroup of Rk . Put B = (1/β)G0 . Then B is invertible, G0 = BX0 with X0 = βI and G1 = BX1 with X1 = X. It follows that Λ(G) is a subset of LB := {BZ : Z ∈ Zk×1 }. As the latter set is discrete, so must be Λ(G). This proves (i). Finally, if β = 1, then B = G0 , which implies that LB ⊂ Λ(G) and hence LB = Λ(G). Consequently, B is a basis matrix for Λ(G). Proposition 3.2 Let G = (G0 G1 ) be a unit equiangular tight (k, n) frame. If Λ(G) is a lattice, then α must be a rational number. Proof. By Proposition 3.1, we may assume that Λ(G) is a full rank lattice. So G = BZ with an invertible matrix B and a matrix Z ∈ Zk×n . Multiplying the equality γI = GG0 = BZZ 0 B 0 from the right by (B 0 )−1 and then from the left by B 0 , we obtain γI = B 0 BZZ 0 and thus, (I + (1/α)C)Z 0 = G0 GZ 0 = Z 0 B 0 BZZ 0 = γZ 0 , which implies that CZ 0 = α(γ − 1)Z 0 . If α is irrational, the last equality yields Z = 0, and this gives G = 0, a contradiction.
4
Unit equiangular tight (k, 2k) frames
√ We first consider the case n = 2k. Then γ = 2 and α = n − 1. We furthermore suppose that n = pr + 1 with an odd prime number p and a natural number r. If r is odd and p = 4` + 3, then k is even, which implies that unit equiangular tight (k, n) frames do not exist (Theorem 17 of [24]). If r is odd and p = 4` + 1, then unit equiangular tight (k, n) frames G exist, but Λ(G) is not a lattice because α is irrational. We are so left with the case where r is even. Theorem 4.1 Let k ≥ 2 and n = 2k. If n = p2m + 1 with an odd prime number p and a natural number m, then there exists a unit equiangular tight (k, n) frame G such that Λ(G) is a full rank lattice. 8
Comments. This theorem proves Theorem 2.2(b) and will be a consequence of the following Theorem 4.2. Before turning to the proof of Theorem 4.2, which is a combination of ideas of Goethals and Seidel [11] and Strohmer and Heath [23], some comments seem to be in order. Following [23], we start with a symmetric n × n conference matrix C, that is, with a symmetric matrix C that has zeros on the main diagonal and ±1 elsewhere and that satisfies C 2 = (n − 1)I. Under the hypothesis of Theorem 4.1, such matrices were first constructed by Paley [16]. Goethals and Seidel [11] showed that one can always obtain such matrices in the form A D C= (3) D −A where A and D are symmetric k × k circulant matrices. Let a and b be any rational numbers such that a2 + b2 = α2 (= n − 1 = p2m ). Theorem 3.4 of [11] says that, under certain conditions, one can in turn represent the matrix (3) as
A D D −A
=
I −N N I
−1
aI bI bI −aI
I −N N I
(4)
with a symmetric circulant matrix N all entries of which are rational numbers. The conditions ensuring the representation (4) are that D + bI or A + aI are invertible. We have N = (A + aI)−1 (bI − D) or N = (D + bI)−1 (A − aI) (5) if A + aI or D + bI is invertible, respectively. (Note that all occurring blocks are symmetric circulant matrices and in particular commuting matrices.) As there are infinitely many different decompositions of p2m into the sum of two squares of rationals, for example, 2 2 2 2ts m t − s2 m 2m p + 2 p p = 2 t + s2 t + s2 with integers s and t, we can, for given A and D, always find rational a and b such that a2 + b2 = α2 and both D + bI and A + aI are invertible. Let, for example n = 10. A matrix (3) with symmetric circulant matrices A and D is completely given by its first line, which is of the form 0, ε1 , ε2 , ε2 , ε1 ,
ε3 , ε4 , ε5 , ε5 , ε4
with εj ∈ {−1, 1} =: {−, +}. These are 25 = 32 matrices. Exactly four of them satisfy C 2 = 9I. Their first lines and the eigenvalues of D are 0, −, +, +, −, 0, −, +, +, −, 0, +, −, −, +, 0, +, −, −, +,
−, +, +, +, +, +, −, −, −, −, −, +, +, +, +, +, −, −, −, −, 9
−2, −2, −2, −2, 3 −3, 2, 2, 2, 2, −2, −2, −2, −2, 3 −3, 2, 2, 2, 2.
(6) (7) (8) (9)
The corresponding matrix A is always singular. We see that in all cases we may take a = 3 and b = 0 (32 + 02 = 9) because D is invertible. In the cases (6) and (8) we could also take a = 0 and b = 3 (02 + 32 = 9) since D + 3I is invertible. In fact, we will prove the following theorem. As shown above, the hypothesis of this theorem can always be satisfied, so that this theorem implies Theorem 4.1. Theorem 4.2 Let k ≥ 2 and n = 2k. Suppose n = p2m + 1 with an odd prime number p and a natural number m, let a and b be rational numbers such that a2 + b2 = p2m and a 6= −pm . Let A and D be symmetric k × k circulant matrices such that (3) is a conference assume A + aI or D + bI is invertible. Define N by (5) and √ matrix, and m put α = n − 1 = p . Then G= p
1 α(α + a)
(I + N 2 )−1/2
(α + a)I + bN bI − (α + a)N
(10)
is a unit equiangular tight (k, n) frame G such that Λ(G) is a full rank lattice. Proof of Theorem 4.2. The requirement a 6= −pm assures that α + a 6= 0. Let 1 W11 W12 U11 U12 2 −1/2 =p W = (I + N ) W21 W22 U21 U22 2α(α + a) with
U11 U12 U21 U22
=
(α + a)I + bN bI − (α + a)N bI − (α + a)N −αI − (α + a)N
.
Using (4) one can show by straightforward computation that U12 U12 U11 U11 , = −α , C =α C U22 U22 U21 U21 which implies that C
U11 U12 U21 U22
=
U11 U12 U21 U22
and thus 2 1/2
C(I + N )
2 1/2
W = (I + N )
W
αI 0 0 −αI αI 0 0 −αI
.
(11)
We have W 2 = I. Indeed, 2 2 U11 + U12 U21 = U21 U12 + U22 = [(α + a)2 + b2 ](I + N 2 ) = 2α(α + a)(I + N 2 ), 2 2 whence W11 +W12 W21 = W21 W12 +W22 = I, and similarly one gets that the off-diagonal 2 blocks of W are zero. From (11) we therefore get αI 0 I 0 αI 0 2 1/2 2 −1/2 C = (I + N ) W W (I + N ) =W W, 0 −αI 0 I 0 −αI
10
or equivalently, 1 I+ C=W α
2I 0 0 0
W =2
2 W11 W12 W11 W21 W11 W21 W12
.
(12)
√ The matrix G given by (10) is just G = 2(W11 W21 ). We claim that G is a unit equiangular tight (k, n) frame. First, since W11 and W21 are symmetric, we have 2 W11 W11 W21 W11 0 W11 W21 = 2 , (13) GG=2 2 W21 W11 W21 W21 and since W21 = W12 , the right-hand sides of (12) and (13) coincide. This proves that G is unit and equiangular. Secondly, W11 2 0 2 ) = 2I, + W21 GG = 2 W11 W21 = 2(W11 W21 which shows that G is tight with √ γ = 2 = n/k. The equality GG0 = 2I implies that the rank of G is k. Thus, G = 2(W11 W12 ) has k linearly independent columns. We permute the columns of G so that these k linearly independent columns become the first k columns. The resulting matrix, which is anew denoted by G, is a unit equiangular tight (k, n) frame of the form G = (G0 G1 ) with an invertible matrix G0 . Furthermore, we have G0 = (I + N 2 )−1/2 R and G1 = (I + N 2 )−1/2 S with matrices R −1 and S whose entries are rational numbers. We therefore obtain that G−1 0 G1 = R S is a matrix with rational entries, and hence, by Proposition 3.1, the set Λ(G) is a full rank lattice. Corollary 4.3 Let k ≥ 2 and n = 2k. Suppose n = p2m + 1 with an odd prime number p and a natural number m, let A and D be symmetric matrices such √ k × k circulant m that the matrix (3) is a conference matrix. Put α = n − 1 = p . If the matrix D is invertible, then I ± (1/α)A are positive definite matrices and, with the invertible matrix N := D−1 (A − αI), √ G := 2(I + N 2 )−1/2 I −N is a unit equiangular tight (k, n) frame G and the set Λ(G) is a full rank lattice. If N ∈ Zk×k , then G may be written as p (14) G = B+ I −N with B+ := I + (1/α)A, and B+ is a basis matrix for Λ(G), while if N −1 ∈ Zk×k , then G may be written in the form p G = B− S −N −1 I with B− := I − (1/α)A, (15) where S := D|D|−1 and |D| is the positive definite square root of D0 D, and this time B− S is a basis matrix for Λ(G), Furthermore, s 1 1 p det B± = det I ± A = k/2 det(αI ± A). α α 11
Remark. Recall that the determinant (= volume of a fundamental domain) of a lattice is defined as the square root of det(B 0 B) where B is any basis p matrix. Thus, if N is an 0 B+ ) = det B+ , integer matrix, then the determinant of the lattice is simply det(B+ −1 while if N has integer entries, the determinant of the lattice Λ(G) is q q q 0 0 0 B− S) = det(SB− B− S) = det(B− B− S 2 ) = det B− det(S 0 B− because S = S 0 and S 2 = I. Proof. Since D is invertible, we may use Theorem 4.2 with a = α and b = 0 (α2 + 02 = α2 ) and with N = D−1 (A − αI). In this special case, formula (10) becomes √ G = 2(I + N 2 )−1/2 I −N , (16) and since N has rational entries,√Proposition 3.1 implies that Λ(G) is a full rank lattice. Proposition 3.1 also shows that 2(I +N 2 )−1/2 is a basis matrix for the lattice provided N ∈ Zk×k . Writing (16) as √ G = − 2(I + N 2 )−1/2 N −N −1 I and permuting (−N −1 I) to (I − N −1 ) we can deduce from Proposition 3.1 that the √ matrix − 2(I + N 2 )−1/2 N is a basis matrix provided N −1 ∈ Zk×k . It remains to show that these two basis matrices are just the matrices B± . As the square of the matrix (3) is α2 I, we have A2 + D2 = α2 I. Since 0 is not in the spectrum of D, the equality A2 + D2 = α2 I implies that the spectrum (= set of eigenvalues) of A is contained in the open interval (−α, α). Hence αI ± A are positive definite. Moreover, we get D2 = α2 I − A2 = (αI − A)(αI + A), and since all involved matrices are circulant and therefore commute, we obtain I + N 2 = I + D−2 (A − αI)2 = I + D−2 (αI − A)(αI − A) = I + (αI − A)−1 (αI + A) = (αI + A)−1 [αI + A + αI − A] = 2α(αI + A)−1 = 2(I + (1/α)A)−1 . √ Consequently, 2(I + N 2 )−1/2 = (I + (1/α)A)1/2 = B+ , which proves (14). The matrix |D| is again a circulant matrix and we have D = S|D| with a circulant matrix S satisfying S 2 = I. From the equality D2 = (αI − A)(αI + A) we obtain that |D| = (αI − A)1/2 (αI + A)1/2 . Thus, √ − 2(I + N 2 )−1/2 N = (I + (1/α)A)1/2 D−1 (αI − A) 1 = √ (αI + A)1/2 (αI − A)−1/2 (αI + A)−1/2 (αI − A)S α 1 = √ (αI − A)1/2 S = (I − (1/α)A)1/2 S = B− S. α This proves (15). The determinant formulas are obvious. 12
Two lattices from (5,10) frames. Let (A, D) be one of the four pairs given by (6) to (9). Thus, k = 5, n = 10, α = 3. In either case, D is invertible det √ D = ±48 √ √with√ and we have det(3I + A) = 48. (The eigenvalues of A are − 5, − 5, 0, 5, 5.) The four circulant matrices N = D−1 (A − 3I) have the first rows (+, 0, −, −, 0),
(+, −, 0, 0, −),
(−, +, 0, 0, +). p Thus, N ∈ Z5×5 , and so by Corollary 4.3, Λ(G) is a lattice, B = I + (1/3)A is a + √ basis matrix, and det B+ = 3−5/2 48 = 22 /32 = 0.4444 . . .. (Incidentally, the matrices N −1 also have integer entries.) The eigenvalues of I + (1/3)A are 1−
1√ 5, 3
(−, 0, +, +, 0),
1−
1√ 5, 3
1,
1+
1√ 5, 3
1+
1√ 5, 3
q √ and hence the smallest eigenvalue of B+ is 1 − (1/3) 5 = 0.50462... > 1/2. We have B+ = U EU 0 with an orthogonal matrix U and the diagonal matrix E of the eigenvalues of B+ . Consequently, 1 1 kB+ xk2 = kU EU 0 xk2 = kEU 0 xk2 > kU 0 xk2 = kxk2 , 4 4 and hence kB+ xk2 > 1 if kxk2 ≥ 4. So consider the x ∈ Z5 \ {0} with kxk2 ≤ 3. Such x contain only 0, +1, −1, and using Matlab we checked that kB+ xk2 < 1.1 for exactly 20 nonzero x of these 35 − 1 = 242 possible x. The 20 columns B+ x are just ± the columns of G. Thus, the minimal distance of Λ(G) is 1, Λ(G) has a basis of minimal vectors, and S(Λ(G)) = {±f1 , . . . , ±f10 }. Note that if we denote the basis matrices for the lattices corresponding to (6) and (9) by B1 , . . . , B4 , then actually B1 = B2 and B3 = B4 . However, B1 B3−1 is not a scalar multiple of an orthogonal matrix. To “see” a concrete matrix B+ , we considered p the case where the matrices A, D are specified by (6). We computed B1 = B+ = I + (1/3)A with Matlab and obtained 0.9303 -0.1651 0.2000 0.2000 -0.1651
-0.1651 0.9303 -0.1651 0.2000 0.2000
0.2000 -0.1651 0.9303 -0.1651 0.2000
0.2000 0.2000 -0.1651 0.9303 -0.1651
-0.1651 0.2000 0.2000 -0.1651 0.9303
√ on the screen. With the Fourier matrix F5 = (1/ 5)(ω (j−1)(k−1) )5j,k=1 , ω = e2πi/5 , this is B1 = F5∗ EF5 with ! r r r r 1√ 1√ 1√ 1√ 1− 5, 1+ 5, 1+ 5, 1− 5 E = diag 1, 3 3 3 3 ! √ √ √ √ 5−1 5+1 5+1 5−1 √ , √ , √ , √ = diag 1, . 6 6 6 6 13
It follows in particular that the numerical values given by Matlab and shown above are p p 0.2000 = 1/5, 0.9303 = 1/5 + 2 2/15, −0.1651 = 1/5 − 2/15. Three lattices from (13,26) frames. Let now k = 13, n = 26, α = 5. Let A and D be symmetric 13 × 13 circulant matrices whose first rows are (0, ε1 , ε2 , ε3 , ε4 , ε5 , ε6 , ε6 , ε5 , ε4 , ε3 , ε2 , ε1 ) and (ε7 , ε8 , ε9 , ε10 , ε11 , ε12 , ε13 , ε13 , ε12 , ε11 , ε10 , ε9 , ε8 ) with εk ∈ {−1, +1} =: {−, +}, respectively. There are 213 = 8 192 such matrices. For exactly 12 of them the matrix (3) satisfies C 2 = 25I. The determinant of D is always det D = ±768 000 = ±212 · 3 · 54 . Thus, by Corollary 4.3, Λ(G) is a full rank lattice. In exactly 6 cases, for example if the first rows of A and D are (0, −, −, −, +, −, +, +, −, +, −, −, −) and (−, −, +, +, +, −, +, +, −, +, +, +, −), p we have N ∈ Z13×13 . We denote the A and B+ = I + (1/5)A corresponding to these cases by A1 , . . . , A6 and B1 , . . . , B6 . Corollary 4.3 implies that Bj is a basis matrix for the jth lattice. We have det(5I + Aj ) = 2 560 000 = 212 · 54 and hence det Bj = 26 /59/2 ≈ 0.0458 for 1 ≤ j ≤ 6. In the other 6 cases, for instance if the first rows of A and D equal (0, −, +, +, +, −, +, +, −, +, +, +, −) and (−, +, −, −, +, +, +, +, +, +, −, −, +), we get that N −1 ∈ Z13×13 . Let Ap 7 , . . . , A12 , S7 , . . . , S12 , and B7 , . . . , B12 be the corresponding A, S = D|D|−1 , B− = I − (1/5)A. We know from Corollary 4.3 that Sj Bj is a basis matrix for the jth lattice. It turns out that det(5I − Aj ) = 2 560 000 = 212 · 54 and hence again det Bj = 26 /59/2 ≈ 0.0458 for 7 ≤ j ≤ 12. Actually, B1 = B2 , B3 = B4 , B5 = B6 , S7 B7 = −S8 B8 , S9 B9 = −S10 B10 , S11 B11 = −S12 B12 , B1 = U1 S11 B11 , B3 = U2 S9 B9 , B5 = U3 S7 B7 with orthogonal matrices U1 , U2 , U3 . The relation “X ∼ Y if and only if XY −1 is a nonzero scalar multiple of an orthogonal matrix” is an equivalence relation on every family of invertible k × k matrices. The equivalence classes of this relation on {B1 , . . . , S12 B12 } are {B1 = B2 , S11 B11 , S12 B12 },
{B3 = B4 , S9 B9 , S10 B10 },
{B5 = B6 , S7 B7 , S8 B8 }.
The first rows of (A1 , D1 ), (A3 , D3 ), (A5 , D5 ) are (0, −, −, −, +, −, +, +, −, +, −, −, −, (0, −, +, +, −, −, −, −, −, −, +, +, −, (0, +, −, −, −, +, −, −, +, −, −, −, +, 14
−, −, +, +, +, −, +, +, −, +, +, +, −), +, −, −, −, +, −, +, +, −, +, −, −, −), +, −, +, +, −, −, −, −, −, −, +, +, −).
For 1 ≤ j ≤ 6, the smallest eigenvalue of Bj is about 0.3736, whence kBj xk2 > 0.372 kxk2 > 0.13kxk2 . Thus, kBxk2 > 1 for kxk2 ≥ 7. In the last 6 cases, the smallest eigenvalue of Bj is about 0.4991 and so we have kSj Bj ck2 = kBj xk2 > 0.492 kxk2 > 0.24kxk2 , which is greater than 1 for kxk2 ≥ 4. We took all j ∈ {1, . . . , 12} and x ∈ Z13 with kxk2 ≤ 6 and checked whether kBj xk2 < 1.1. For each j, we obtained exactly 52 vectors x ∈ Z13 \ {0} such that kBj xk2 < 1.1. The columns Bj x are ± the 26 columns f1 , . . . , f26 of G. Consequently, in all cases the minimal distance of Λ(G) is 1, Λ(G) has a basis of minimal vectors, and S(Λ(G)) = {±f1 , . . . , ±f26 }. Ten lattices from (25,50) frames. We finally take k = 25, n = 50, α = 7. We consider the 25 × 25 circulant matrices A and D whose first rows are (0, ε1 , ε2 , ε3 , ε4 , ε5 , ε6 , ε7 , ε8 , ε9 , ε10 , ε11 , ε12 , ε12 , ε11 , ε10 , ε9 , ε8 , ε7 , ε6 , ε5 , ε4 , ε3 , ε2 , ε1 ) and (ε25 , ε13 , ε14 , ε15 , ε16 , ε17 , ε18 , ε19 , ε20 , ε21 , ε22 , ε23 , ε24 , ε24 , ε23 , ε22 , ε21 , ε20 , ε19 , ε18 , ε17 , ε16 , ε15 , ε14 , ε13 ), with εk ∈ {−1, 1} =: {−, +}, respectively. These are 225 = 33 554 432 matrices. In exactly 20 cases the matrix C given by (3) satisfies C 2 = 49 I. One such case is where the first rows of A and D are (0, −, −, −, +, −, +, +, −, +, +, +, −, −, +, +, +, −, +, +, −, +, −, −, −) and (−, −, +, +, +, +, +, −, +, −, +, +, −, −, +, +, −, +, −, +, +, +, +, +, −), respectively. We have | det D| = det(7I + A) = det(7I − A) = 260 119 8402 = 222 · 32 · 52 · 72 · 114 , N ∈ Z25×25 and N −1 ∈ Z25×25 in each p of the 20 cases. Thus, by Corollary 4.3, we obtain 20 lattices Λ(Gj ) with Bj = I + (1/7)Aj as a basis matrix and det Bj =
211 · 3 · 5 · 7 · 112 211 · 3 · 5 · 112 = ≈ 0.0007 1052 725/2 723/2
for all 1 ≤ j ≤ 20. In fact Bj = Bj+10 for 1 ≤ j ≤ 10, and the equivalence classes of the set {B1 , . . . , B10 } under the equivalence relation “Bi ∼ Bj if and only if Bi Bj−1 is a nonzero scalar multiple of an orthogonal matrix” are the ten singletons {B1 }, . . . , {B10 }. The smallest eigenvalue of Bj is about 0.1415 for all j. 15
5
Unit equiangular tight (k, k + 1) frames
Sometimes it is advantageous to represent a unit equiangular tight (k, n) frame by coordinates different from those in Rk . This is in particular the case for (k, k + 1) frames. Fix k ≥ 2 and consider the set F of the k + 1 normalized columns of height k + 1 formed by the permutations of −k, 1, . . . , 1, −k 1 1 1 −k 1 1 1 1 f1 = √ .. , f2 = √ 2 .. , . . . , fk+1 = √ 2 .. . 2 k +k . k +k . k +k . 1 1 −k These k + 1 vectors are in the orthogonal complement of (1, . . . , 1)0 ∈ Rk+1 and may therefore be thought of as vectors in Rk . Let Λ(F) = spanZ {f1 , . . . , fk+1 } ⊂ Rk . The following theorem in conjunction with Proposition 2.4 proves Theorem 2.2(a). Theorem 5.1 The vectors f1 , . . . , fk+1 form a unit equiangular tight (k, k + 1) frame and Λ(F) is a full rank lattice. The matrix B constituted by f1 , . . . , fk , −k 1 ... 1 1 −k . . . 1 1 .. . . . . , (17) B=√ . . . 2 k +k 1 1 . . . −k 1 1 ... 1 (k+1)×k is a basis matrix for Λ(F), we have 1 det(B B) = k+1 0
k 1 1+ , k
the lattice Λ(F) has a basis of minimal vectors, and S(Λ(F)) = {±f1 , . . . , ±fk+1 }. Proof. It is well known that F is a unit equiangular tight (k, k + 1) frame. We include the proof for the reader’s convenience. First, the columns of the matrix B are easily seen to be linearly independent, which shows that spanR {f1 , . . . , fk } = Rk . Secondly, it is clear that kfj k = 1 for all j. Thirdly, we have (fi , fj ) = (−k − 1)/(k 2 + k) = −1/k for i 6= j. And finally, a basic result by Welch [28], proofs of which p are also in [6, 23, 24], says that if we are given a unit equiangular frame with |(fi , fj )| p = (n − k)/(k(n − 1)) for i 6= j, then (and only then) the frame is tight. In our case (n − k)/(k(n − 1)) = 1/k, so that the equality is satisfied, implying the tightness of the frame. Incidentally, a 16
completely elementary argument is as follows. If x = (x1 , . . . , xk+1 ) and x1 +· · ·+xk+1 = 0, then ! X 1 1 −kxj + (−kxj − xj ) (fj , x) = √ xi = √ 2+k k2 + k k i6=j and hence
k+1 X
k+1 1 X k+1 (fj , x) = 2 (−(k + 1)xj )2 = kxk2 , k + k j=1 k j=1 2
that is, the frame is tight with γ = (k + 1)/k. Since f1 + · · · + fk = −fk+1 , we have Λ(F) = spanZ {f1 , . . . , fk }. This shows that Λ(F) is {BX : X ∈ Zk }. Consequently, Λ(F) is a full rank lattice with the matrix B given by (17) as a basis matrix. The product B 0 B is a b ... b 1 b a ... b 0 (18) BB= 2 . . . k + k .. .. . . . .. b b . . . a k×k with a = k 2 + k and b = −k − 1. The determinant of a matrix of the form (18) is known to be (a − b)k−1 (a + (k − 1)b). Thus, det B 0 B =
1 (k + 1)k−1 2 k−1 2 (k + k + k + 1) (k + k − (k − 1)(k + 1)) = . (k 2 + k)k kk
We are left with determining S(Λ(F)). Straightforward computation shows that the inequality kBxk2 ≥ 1 is equivalent to the inequality (k + 1)(x21 + · · · + x2k ) ≥ k + (x1 + · · · + xk )2 ,
(19)
and that equality holds in both inequalities only simultaneously. We first show (19) for integers (x1 , . . . , xk ) ∈ Zk \ {0} by induction on k. For k = 1, inequality (19) is trivial. Suppose it is true for k − 1: k(x21 + · · · + x2k−1 ) ≥ k − 1 + (x1 + · · · + xk−1 )2 . If x21 + · · · + x2k−1 ≥ 1, we may add x21 + · · · + x2k−1 on the left and 1 on the right to get (k + 1)(x21 + · · · + x2k−1 ) ≥ k + (x1 + · · · + xk−1 )2 . This proves (19) in the case where one of the integers x1 , . . . , xk is zero and one of them is nonzero. We are so left with the case where xj 6= 0 for all j. Then x21 + · · · + x2k ≥ k and hence k + (x1 + · · · + xk )2 ≤ k + (|x1 | + · · · + |xk |)2 ≤ k + k(x21 + · · · + x2k ) ≤ x21 + · · · + x2k + k(x21 + · · · + x2k ) = (k + 1)(x21 + · · · + x2k ), 17
(20) (21)
which completes the proof of (19). At this point we have shown that {f1 , . . . , fk } is a basis of minimal vectors. To identify all of S(Λ(F)), we have to check when equality in (19) holds. Suppose first that xj 6= 0 for all j. In that case we have (20) to (21). Equality in (21) holds if and only if |xj | = 1 for all j, and equality in (20) is valid if and only if all the xj have the same sign. Thus, we get the two vectors x = (1, . . . , 1)0 and x = (−1, . . . , −1)0 . The corresponding products Bx are −fk+1 and fk+1 . Suppose finally that one of the xj is zero, say xk = 0. From (19) with k replaced by k − 1 we know that k(x21 + · · · + x2k−1 ) ≥ k − 1 + (x1 + · · · + xk−1 )2 . If x21 + · · · + x2k−1 > 1, we may add this inequality to the previous one to obtain that (k + 1)(x21 + · · · + x2k−1 ) ≥ k + (x1 + · · · + xk−1 )2 . Consequently, for x21 +· · ·+x2k−1 > 1 equality in (19) does not hold. If x21 +· · ·+x2k−1 = 1, then xj = ±1 for some j and xi = 0 for all i 6= j. In that case equality in (19) holds and the vector Bx is ±fj . In summary, we have proved that the set S(Λ(F)) of all minimal vectors is just {±f1 , . . . , ±fk+1 }.
6
The remaining frames in dimensions at most 9
Recall that (2) lists the unit equiangular tight frames in dimensions k ≤ 9 different from the (k, k + 1) frames. By Proposition 2.1, the (3, 6), (7, 14), and (9, 18) frames do not yield lattices, and the lattices resulting from the (5, 10) case were discussed in Section 4. We are left with the (6, 16) and (7, 28) cases. A lattice from a frame + + 1 + G= √ 6 + + +
(6,16) frame. In [24] we see the unit equiangular tight (6, 16) + + + + − −
+ + + − + −
+ + + − − +
+ + − + + −
+ + − + − +
+ + − − + +
+ + − − − −
+ − + + + −
+ − + + − +
+ − + − + +
+ − + − − −
+ − − + + +
+ − − + − −
+ − − − + −
+ − − − − +
.
Here GG0 = (16/6)I and G0 G = I +(1/3)C with a 16×16 matrix C whose diagonal entries are zero and the other entries of which are ±1. The six columns f1 , f2 , f3 , f4 , f5 , f9 of the matrix G are linearly independent and each of the remaining 10 columns is a linear combination with integer coefficients of these six columns. Consequently, by
18
Proposition 3.1 with β = 1, these six columns form a basis matrix, + + + + + + + + + + + − 1 + + + + − + . B=√ 6 + + − − + + + − + − + + + − − + − − We have det(B 0 B) = 26 /36 . With B 0 B = U 0 EU , we get kBxk2 = (EU x, U x) ≥ 0.48kxk2 , and this is at least 6 if kxk2 ≥ 13. So consider the x ∈ Z6 \ {0} with kxk2 ≤ 13. Such x contain only 0, ±1, ±2, ±3, and using Matlab we checked that kBxk2 < 6.1 for exactly 32 nonzero 6 x √ of these 7 − 1 = 117 648 possible x. The 32 columns Bx are just ± the columns of 6G. Thus, Λ(F) has a basis of minimal vectors and S(Λ(G)) = {±f1 , . . . , ±f16 }. A perfect lattice from a (7,28) frame. It is well known that the 82 = 28 vectors resulting from the columns (−3, −3, 1, 1, 1, 1, 1, 1)0 by permuting the entries form an equiangular tight (7, 28) frame. To be precise, let F be the set of the vectors −3 1 −3 1 1 1 1 1 1 1 . f1 = √ , . . . , f28 = √ 1 1 24 24 1 1 1 −3 1 −3 These are unit vectors in R8 . They are all orthogonal to the vector (1, 1, 1, 1, 1, 1, 1, 1)0 , and after identifying the orthogonal complement of this vector with R7 , we may think of f1 , . . . , f28 as unit vectors in R7 . We consider the set Λ(F) = spanZ {f1 , . . . , f28 } ⊂ R7 . The columns of the 8 × 7 matrix −3 −3 −3 −3 −3 −3 1 −3 1 1 1 1 1 1 1 −3 1 1 1 1 −3 1 1 1 −3 1 1 1 1 B=√ 1 1 −3 1 1 1 24 1 1 1 1 1 −3 1 −3 1 1 1 1 1 −3 1 1 1 1 1 1 1 1 19
are formed by 7 of the above 28 vectors. We denote these 7 vectors by f1 , . . . , f7 . For the reader’s convenience, we show that {f1 , . . . , f28 } is a unit equiangular tight (7, 28) frame. The rank of the matrix B is 7, and hence spanR {f1 , . . . , f28 } = R7 . Clearly, kfj k = 1 for all j.pWe have |(fi , fj )| = 8/24 = 1/3 for i 6= j, which gives the equiangularity, and since (n − k)/(k(n − 1)) = 1/3, we deduce from Welch’s result cited in the proof of Theorem 5.1 that the frame is tight. Straightforward inspection shows that each of the vectors f8 , . . . , f28 is a linear combination with integer coefficients of the vectors f1 , . . . , f7 . Consequently, Λ(F) is a full rank lattice in R7 , {f1 , . . . , f7 } is a basis of Λ(F), and B is a basis matrix. We have 24 8 8 8 8 8 −8 8 24 8 8 8 8 8 8 8 24 8 8 8 −8 1 0 8 8 8 24 8 8 −8 BB= , 24 8 8 24 8 8 8 8 8 8 8 8 8 24 −8 −8 8 −8 −8 8 −8 24 and straightforward computation gives det B 0 B =
26 227 = . 247 37
We now prove that the minimal norm of Λ(F) is 1. Let −3 −3 −3 −3 −3 −3 1 −3 1 1 1 1 1 1 1 −3 1 1 1 1 −3 √ 1 1 −3 1 1 1 1 e = 24B = . B 1 1 1 −3 1 1 1 1 1 1 1 −3 1 −3 1 1 1 1 1 −3 1 1 1 1 1 1 1 1 e ∈ Z8 . We are interested in the x for which kyk2 ≤ 24. Take x ∈ Z7 and consider y = Bx With s := x1 + · · · + x7 , we have y1 = −3s + 4x7 , y3 = s − 4x2 − 4x7 , y6 = s − 4x5 − 4x7 , y8 = s, y2 = s − 4x1 , y4 = s − 4x3 , y5 = s − 4x4 , y7 = s − 4x6 . It suffices to search for all x ∈ Z7 with s ≥ 0 and y12 + · · · + y82 ≤ 24. This is impossible for y8 = s > 5. So we may assume that 0 ≤ s ≤ 4. Suppose first that s = 4. We then must have y12 + · · · + y72 ≤ 9. Since y1 is an even number, it cannot be ±3. Consequently, −2 ≤ −3s + 4x7 = −12 + 4x7 ≤ 2, which gives x7 = 3. Analogously, as y3 is even, we get −2 ≤ s − 4x2 − 4x7 = −8 − 4x2 ≤ 2, 20
which yields x2 = −2. In the same way we obtain x5 = −2. Finally, the even number s − 4x1 = 4 − 4x1 is at least −2, which implies that x1 ≤ 1. Equally, x3 , x4 , x6 ≤ 1. It follows that s = x1 + · · · + x7 ≤ 1 + 1 + 1 + 1 − 2 − 2 + 3 = 3 < 4 = s, which is a contradiction. Thus, we may restrict our search to 0 ≤ s ≤ 3 and y12 + · · · + y72 ≤ 24. The inequality −4 ≤ −3s + 4x7 ≤ 4 gives −4 ≤ 3s − 4 ≤ 4x7 ≤ 3s + 4 ≤ 13, whence −1 ≤ x7 ≤ 3. These are 5 possibilities. From −4 ≤ s − 4xj ≤ 4 we obtain that −1 ≤ xj ≤ 1 for j = 1, 3, 4, 6, which is 34 possibilities, and the inequality −4 ≤ s − 4xj − 4x7 ≤ 4 delivers −16 ≤ s − 4 + 4x7 ≤ 4yj ≤ 4 + s − 4x7 ≤ 11 and hence −4 ≤ xj ≤ 2 for j = 2, 5, leaving us with 72 possibilities. In summary, we have to check 5 · 34 · 72 = 19 845 possibilities. Matlab does this with integer arithmetics within a second. The result is that 0 ≤ s ≤ 3 and y12 + · · · + y72 ≤ 24 happens in exactly 50 cases. One of these √ cases is y = 0, and in the remaining 49 cases y is ± one of the 2 · 28 = 56 vectors 24fj . (Recall that, by symmetry, we restricted ourselves to s ≥ 0. For −3 ≤ s ≤ 3 and y12 + · · · + y72 ≤ 24 to happen √ we would obtain exactly 57 cases: the case y = 0 and the 56 vectors y given by ± 24fj .) This proves that the minimal distance of Λ(G) is 1, that S(Λ(G)) = {±f1 , . . . , ±f28 }, and that Λ(G) has a basis of minimal vectors. From Proposition 2.4 we deduce that the lattice Λ(G) is strongly eutactic. We finally show that this (7, 28) frame generates a perfect lattice. We have shown that the 28 lattice vectors f1 , . . . , f28 are minimal vectors. These vectors are given by their coordinates in the ambient R8 . We use a special 7 × 8 matrix A to transform these vectors isometrically into R7 . The jth row of A is 1 p (1, . . . , 1, −j, 0, . . . , 0) j2 + j p with j ones and 7 − j zeros. We have A = EA0 with E = diag(1/ j 2 + j)7j=1 and with (1, . . . , 1, −j, 0, . . . , 0) being the jth row of A0 . We then get the 28 minimal vectors Afj = EA0 fj (j = 1, . . . , 28) in R7 . These give us 28 symmetric 7 × 7 matrices Cj = E(A0 fj )(A0 fj )0 E. The lattice Λ(F) is perfect if the real span of these 28 matrices is the space of all 7 × 7 symmetric matrices. Each symmetric 28 × 28 matrix may be written as ET E with a symmetric matrix T , and hence we are left with showing that each symmetric 28×28 symmetric matrix T is a real linear combination of the matrices (A0 fj )(A0 fj )0 . For k = 1, . . . , 7, let ([Cj ]k,k , [Cj ]k+1,k , . . . , [Cj ]7,k )0 21
be the column formed by the entries of the kth column of Cj that are on or below the main diagonal. Stack these columns to a column Dj of height 7 + 6 + · · · + 1 = 28. The lattice is perfect if and only if the real span of D1 , . . . , D28 is all of R28 , which happens if and only if the 28 × 28 matrix D constituted by the 28 columns D1 , . . . , D28 is invertible. Tables 2 and 3 show the matrix D. Table 2: The first 14 columns of the matrix D.
0 0 0 0 0 0 0 49 42 35 28 21 14 36 30 24 18 12 25 20 15 10 16 12 8 9 6 4
16 −4 24 20 16 12 8 1 −6 −5 −4 −3 −2 36 30 24 18 12 25 20 15 10 16 12 8 9 6 4
16 12 −8 20 16 12 8 9 −6 15 12 9 6 4 −10 −8 −6 −4 25 20 15 10 16 12 8 9 6 4
16 12 8 −12 16 12 8 9 6 −9 12 9 6 4 −6 8 6 4 9 −12 −9 −6 16 12 8 9 6 4
16 12 8 4 −16 12 8 9 6 3 −12 9 6 4 2 −8 6 4 1 −4 3 2 16 −12 −8 9 6 4
16 12 8 4 0 −20 8 9 6 3 0 −15 6 4 2 0 −10 4 1 0 −5 2 0 0 0 25 −10 4
16 12 8 4 0 −4 −24 9 6 3 0 −3 −18 4 2 0 −2 −12 1 0 −1 −6 0 0 0 1 6 36
16 4 −24 −20 −16 −12 −8 1 −6 −5 −4 −3 −2 36 30 24 18 12 25 20 15 10 16 12 8 9 6 4
16 −12 8 −20 −16 −12 −8 9 −6 15 12 9 6 4 −10 −8 −6 −4 25 20 15 10 16 12 8 9 6 4
16 −12 −8 12 −16 −12 −8 9 6 −9 12 9 6 4 −6 8 6 4 9 −12 −9 −6 16 12 8 9 6 4
16 −12 −8 −4 16 −12 −8 9 6 3 −12 9 6 4 2 −8 6 4 1 −4 3 2 16 −12 −8 9 6 4
16 −12 −8 −4 0 20 −8 9 6 3 0 −15 6 4 2 0 −10 4 1 0 −5 2 0 0 0 25 −10 4
16 −12 −8 −4 0 4 24 9 6 3 0 −3 −18 4 2 0 −2 −12 1 0 −1 −6 0 0 0 1 6 36
0 0 0 0 0 0 0 25 10 −25 −20 −15 −10 4 −10 −8 −6 −4 25 20 15 10 16 12 8 9 6 4
The matrix D can be constructed with integer arithmetics. The determinant det D may be computed by the Gaussian algorithm and thus with integer arithmetics, too. In the intermediate steps, one may factor out powers of 2. For example, in the original matrix D we may draw out 16 from the first line, 4 from the second, 8 from the third, and so on. It results that e = 248 det D, e det D = 163 · 84 · 49 · 26 · det D e The final result is and we may start the Gaussian algorithm with det D. det D = 3 · 2159 . As this is nonzero, we conclude that D is invertible and thus that Λ(F) is perfect. At this point the proof of Theorem 2.2(c) is complete. The perfection of this lattice was also established by Bacher in [2] (see Section 7, especially 7.1). However, Bacher’s approach is different from ours: he obtains the lattice in question as the kernel of a certain linear map, establishes its perfection, and 22
Table 3: The last 14 columns of the matrix D.
0 0 0 0 0 0 0 25 −10 15 −20 −15 −10 4 −6 8 6 4 9 −12 −9 −6 16 12 8 9 6 4
0 0 0 0 0 0 0 25 −10 −5 20 −15 −10 4 2 −8 6 4 1 −4 3 2 16 −12 −8 9 6 4
0 0 0 0 0 0 0 25 −10 −5 0 25 −10 4 2 0 −10 4 1 0 −5 2 0 0 0 25 −10 4
0 0 0 0 0 0 0 25 −10 −5 0 5 30 4 2 0 −2 −12 1 0 −1 −6 0 0 0 1 6 36
0 0 0 0 0 0 0 1 6 3 −4 −3 −2 36 18 −24 −18 −12 9 −12 −9 −6 16 12 8 9 6 4
0 0 0 0 0 0 0 1 6 −1 4 −3 −2 36 −6 24 −18 −12 1 −4 3 2 16 −12 −8 9 6 4
0 0 0 0 0 0 0 1 6 −1 0 5 −2 36 −6 0 30 −12 1 0 −5 2 0 0 0 25 −10 4
0 0 0 0 0 0 0 1 6 −1 0 1 6 36 −6 0 6 36 1 0 −1 −6 0 0 0 1 6 36
0 0 0 0 0 0 0 1 2 7 4 −3 −2 4 14 8 −6 −4 49 28 −21 −14 16 −12 −8 9 6 4
0 0 0 0 0 0 0 1 2 7 0 5 −2 4 14 0 10 −4 49 0 35 −14 0 0 0 25 −10 4
0 0 0 0 0 0 0 1 2 7 0 1 6 4 14 0 2 12 49 0 7 42 0 0 0 1 6 36
0 0 0 0 0 0 0 1 2 3 8 5 −2 4 6 16 10 −4 9 24 15 −6 64 40 −16 25 −10 4
0 0 0 0 0 0 0 1 2 3 8 1 6 4 6 16 2 12 9 24 3 18 64 8 48 1 6 36
0 0 0 0 0 0 0 1 2 3 4 9 6 4 6 8 18 12 9 12 27 18 16 36 24 81 54 36
then remarks that its set of minimal vectors comprises an equiangular system. We, on the other hand, construct the lattice from the equiangular frame and show its perfection directly from this construction. Hence our argument here complements Bacher’s, going in the opposite direction. Since Λ(F) is perfect and strongly eutactic, the packing density of this lattice is a local maximum. As we know the minimal distance and the determinant of this lattice, the packing density can be easily computed using (1). It turns out to be 21.57 %. This is better than the packing density of the root lattice A7 , which is 14.76 %. In [4], we studied lattices in Rk that are generated by Abelian groups of the order k + 1. There the packing density of the lattices generated by Abelian groups of order 8 was shown to 20.88 %. Thus, Λ(F) is also better than this. We nevertheless do not reach the best packing density for a 7-dimensional lattice, which is 29.53 % and is achieved for the well known lattice E7 . Acknowledgements. Fukshansky acknowledges support of the NSA grant H982301510051. Garcia acknowledges support of the NSF grant DMS-1265973. Needell acknowledges support of the Alfred P. Sloan Fellowship and NSF Career grant number 1348721. We sincerely thank the two referees for their competent comments and valuable suggestions.
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A. B¨ottcher, Fakult¨at f¨ ur Mathematik, TU Chemnitz, 09107 Chemnitz, Germany
[email protected] L. Fukshansky, Department of Mathematics, Claremont McKenna College, 850 Columbia Ave, Claremont, CA 91711, USA
[email protected] S. R. Garcia, Department of Mathematics, Pomona College, 610 N. College Ave, Claremont, CA 91711, USA 25
[email protected] H. Maharaj, Department of Mathematics, Pomona College, 610 N. College Ave, Claremont, CA 91711, USA
[email protected] D. Needell, Department of Mathematics, Claremont McKenna College, 850 Columbia Ave, Claremont, CA 91711, USA
[email protected]
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